Problem: Divide the following complex numbers. $ \dfrac{-18-12i}{1+5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-5i}$ $ \dfrac{-18-12i}{1+5i} = \dfrac{-18-12i}{1+5i} \cdot \dfrac{{1-5i}}{{1-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-18-12i) \cdot (1-5i)} {(1+5i) \cdot (1-5i)} = \dfrac{(-18-12i) \cdot (1-5i)} {1^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-18-12i) \cdot (1-5i)} {(1)^2 - (5i)^2} = $ $ \dfrac{(-18-12i) \cdot (1-5i)} {1 + 25} = $ $ \dfrac{(-18-12i) \cdot (1-5i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-18-12i}) \cdot ({1-5i})} {26} = $ $ \dfrac{{-18} \cdot {1} + {-12} \cdot {1 i} + {-18} \cdot {-5 i} + {-12} \cdot {-5 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{-18 - 12i + 90i + 60 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{-18 - 12i + 90i - 60} {26} = \dfrac{-78 + 78i} {26} = -3+3i $